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Let $P$ be a chain-complete poset with a least element, and let $f_1,f_2,\ldots,f_n$ be order-preserving maps $P\to P$ such that $\forall i,j: f_i \circ f_j = f_j\circ f_i$.

Claim. The functions $f_1,\ldots,f_n$ have a common fixed point.

Proof (incomplete). By this question we know $P_{f_i}=\{x\in P|f_i(x)=x\}$ is chain-complete with a least element for all $i$. Now for any $i,j$ let $x$ be a fixed point of $f_i$. Then $$(f_i\circ f_j)(x)=(f_j\circ f_i)(x) = f_j(x)$$ so $f_j(x)$ is also a fixed point of $f_i$. Thus $f_j\restriction P_{f_i}:P_{f_i}\to P_{f_i}$ so $f_j$ has a least fixed point on $P_{f_i}$.

This covers the case $n=2$. Can I use induction from here?

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Yes I can. By induction, the set of common fixed points of $f_1, \ldots, f_{n-1}$ is chain-complete and has a least element. But then again $f_n$ has a least fixed point if restricted to this set.

I struggled because I tried "have common fixed point" as the induction hypotheses. Taking the stronger hypotheses "the set of common fixed points is chain-complete and has a least element" makes the induction easy.

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